Which method can you use to prove these triangles congruent ?

a- the sas postulate
b-the hl theorem
c-the sss postulate
d- aas theorem

plz m im almost failing

The HL theorem ⇒ answer (b)

Step-by-step explanation:

* Lets revise the cases of congruence

- SSS  ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ

- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and

 including angle in the 2nd Δ

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ

 ≅ 2 angles and the side whose joining them in the 2nd Δ

- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles

 and one side in the 2ndΔ

- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse

 leg of the 2nd right angle Δ

* In the problem

- In Δ ABC

∵ AC = 10 units

∵ BC = 6 units

∵ m∠ABC = 90°

- In Δ DEF

∵ DF = 10 units

∵ EF = 6 units

∵ m∠DEF = 90°

∴ AC = DF = 10 units

∵ BC = EF = 6 units

∵ m∠∠B = m∠E = 90°

∴ The two triangles ABC and DEF are congruent by using

  the HL theorem (hypotenuse leg of a right triangle)

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?